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Topic: Titration of weak base and strong acid  (Read 3858 times)

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Offline neon

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Titration of weak base and strong acid
« on: April 16, 2014, 01:56:15 PM »
A 2.5 gm impure sample containing weak monoacidic base (Mol wt =45) is dissolved in 100 ml water and titrated with 0.5M HCl when 1/5 of the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Find concentration of salt at equivalence point.

Let the initial concentration of base be c. When 1/5 of the base is neutralised amount of base left is equal to 4c/5 and concentration of salt formed is c/5.
Using Henderson's equation
pOH = pKb +log(1/4)

This gives me pKb = 5.6.
At equivalence point
Total amount of base is converted into salt. Thus concentration of salt is c.
Assuming hydrolysis of the salt
pH = 7-1/2pKb-log(c)

This gives me C=0.5. But the correct answer says that it should be 0.25.

Offline Borek

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Re: Titration of weak base and strong acid
« Reply #1 on: April 16, 2014, 04:14:40 PM »
pH = 7-1/2pKb-log(c)

Check it out.
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Offline neon

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Re: Titration of weak base and strong acid
« Reply #2 on: April 16, 2014, 10:44:01 PM »
What's wrong with that equation? In salt hydrolysis of a salt containing weak base and strong acid,

[itex][H+] = \sqrt{\dfrac{K_w C}{K_b}} [/itex]

Converting the above into logarithms gives exactly the same expression I wrote earlier.

Offline Borek

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Re: Titration of weak base and strong acid
« Reply #3 on: April 17, 2014, 03:19:38 AM »
Converting the above into logarithms gives exactly the same expression I wrote earlier.

That's where you are wrong. Check your math.
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Offline neon

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Re: Titration of weak base and strong acid
« Reply #4 on: April 17, 2014, 03:39:17 AM »
Oops! That was a silly mistake.  :P Thanks for pointing out.

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